An alternating voltage of $22 \mathrm{~V}$ is applied across an inductance coil of reactance $11 \Omega$. If…
An alternating voltage of $22 \mathrm{~V}$ is applied across an inductance coil of reactance $11 \Omega$. If the inductance of the coil is $0.07 \mathrm{H}$, the frequency of alternating voltage is
$25 \mathrm{~Hz}$
$40 \mathrm{~Hz}$
$50 \mathrm{~Hz}$
$20 \mathrm{~Hz}$
Solution
The inductive reactance is
$\begin{aligned}
& X_L=\omega L \\
& \omega=2 \pi f \\
& \therefore 2 \pi f L=X_L \\
& \Rightarrow f=\frac{X_L}{2 \pi L}=\frac{11}{2 \pi(0.07)} \mathrm{Hz}=25 \mathrm{~Hz}
\end{aligned}$